**Physics 1151
Exam #3
November 15, 2002**

**Possibly useful equations:**

My powers of intuition and guessing are non-existent. __Please__
make your answers clear and easy to follow. Unlabeled numbers make it very difficult
to guess or figure out what you are doing. Tell me the details! Label and explain
what you are doing.

**1.** (9.29) A solid, uniform
cylinder of 12 cm radius with mass of 5.0 kg is free to rotate about its symmetry
axis. A cord is wound around the drum and a 1.2 kg mass is attached to the end
of the cord. Find the acceleration of the hanging mass, the angular acceleration
of the cylinder, and the tension in the cord.

For the 1.2 kg mass, take down as positive and apply Newton's Second Law,

F_{net} = m a

F_{net} = 12 N - T = (1.2 kg) a

12 N - T = (1.2 kg) a

For the rotating cylinder, apply the rotational form of Newton's Second Law,

_{net} = I

_{net}
= r F sin

_{net} = (0.12
m) (T) (1)

since = 90°

For a solid cylinder,

I = (1/2) M r^{2}

I = (0.5) (5.0 kg) (0.12 m)^{2}

I = 0.036 kg m^{2}

Therefore,

_{net} = (0.12
m) T = (0.036 kg m^{2})

At the moment, we have three unknowns, T, a, and ,
in two equations. The angular acceleration of the cylinder and the linear acceleration
of the cord or of the hanging mass are connected by

a = r

a = (0.12 m)

= a / 0.12 m

(0.12 m) T = (0.036 kg m^{2})

(0.12 m) T = (0.036 kg m^{2}) [ a / 0.12 m]

T = (0.036 kg m2) a/[ (0.12 m)^{2}(0.12 m) ]

T = (0.036 kg m^{2}) a/(0.12 m)^{2}

T = (2.5 kg) a

Now we have two equations in two unknowns

12 N - T = (1.2 kg) a

12 N - (2.5 kg) a = (1.2 kg) a

12 N - (2.5 kg) a = (1.2 kg) a

12 N = (3.7 kg) a

a = 12 N / 3.2 kg

a = 3.24 m/s^{2}

T = 8.1 N

**2.** (10.56) What percentage of a copper cube floats above
the surface if it floats in mercury?

(Hg)
= 13,600 kg/m^{3}

(Cu)
= 8,920 kg/m^{3}

We can immediately apply Bernoulii’s Equation,

or

In our case, the hydrant is position 1 and the top of the stream of water is position 2,

v

h

p

and

v

h

p

(The guage pressure at the top of the stream is zero).

p

p

p

p

p

Remember, the subscript "g" simply means "gauge pressure".

**4.** (12.57) A 0.050-kg
ice cube, initially at – 5.0°C, is placed in 0.30 kg of water at 25°C.

What is the final temperature of the water?

Or, if melting is not complete, how much ice remains unmelted at thermal equilibrium?

Assume no heat is lost to the environment.

First, assume all the ice melts, so T_{f} > 0°C

Q_{1} + Q_{2} = 0

Q_{1} = heat lost by the water

Q_{1} = c_{water} m_{water}T_{water}
= (4186 ^{J}/_{kg C°})(0.30 kg)(T_{f} – 25°C)

Q_{1} = (4186 ^{J}/_{kg C°})(0.30 kg)(T_{f}
– 25°C) = 1,256 (^{J}/_{C°}) T_{f} –
31,395 J

Q_{2} = heat gained by ice

Q_{2} = heat gained in raising ice to 0°C +

+ heat gained in melting ice +

+ heat gained in raising the melted ice (now water) to temperature T_{f}

Q_{2} = c_{ice} m_{ice} T_{1}
+ L_{f,ice} m_{ice} + c_{water} m_{ice} DT_{2}

Q_{2} = [2090 ^{J}/_{kg C°}][0.05 kg][0°C –
(– 5°C)] +

+ [3.33 x 10^{5} ^{J}/_{kg}][0.05 kg] +

+ [4186 ^{J}/_{kg C°}][0.05 kg][T_{f} – 0°C]

__Be very careful with these temperature changes !__

Q_{2} = 522.5 J + 16,650 J + 209.3 (^{J}/_{C°})
T_{f} – 0

Q_{2} = 17,172.5 J + 209.3 (^{J}/_{C°}) T_{f}

Now we are ready to go back to

Q_{1} + Q_{2} = 0

1,256 (^{J}/_{C°}) T_{f} – 31,395 J + 17,172.5
J + 209.3 (^{J}/_{C°}) T_{f} = 0

1,256 (^{J}/_{C°}) T_{f} + 209.3 (^{J}/_{C°})
T_{f } = 31,395 J – 17,172.5 J

1,465 (^{J}/_{C°}) T_{f} = 14,222 J

T_{f} = [^{14,222}/_{1,465}]°C

T_{f} = 9.7°C

**5. Concept Questions:
i)** Consider a twirlling ice skater (or a dizzy Physics professor) who spends
faster as she (or he) pulls her (or his) hands in close. Spinning increases
because angular momentum is conserved. However, the rotational

**As a liquid is heated, its volume expands. This can be used
to make a common liquid-in-glass thermometer.
As a metal is heated, its length increases. This can be used to make a bimetallic
strip thermometer.
As a resistor is heated, its resistance changes. This can be used to make an
electronic thermometer.
As a gas is heated, its volume expands. This can be used to make a constant
volume gas thermometer. We used such a thermometer to come up with the absolute
temperature scale, now more usually known as the Kelvin temperature scale. **